Q.Use of hot air balloon in sports and meteorological observation is an application of

At constant room temperature, Boyle's Law (P1V1 = P2V2) applies. Using the given initial condition (1 bar, 2.27 L) and the bursting pressure (0.2 bar), the maximum volume works out to 11.35 L.

Since the balloon is at constant (room) temperature throughout, and the amount of gas inside doesn't change, this is a direct application of Boyle's Law:

$P_1 V_1 = P_2 V_2$

Given:

• $P_1 = 1\ \text{bar}$, $V_1 = 2.27\ \text{L}$ (the stated initial condition)
• $P_2 = 0.2\ \text{bar}$ (the pressure at which the balloon bursts — this is the lowest pressure the balloon can withstand before failing, which corresponds to its maximum safe volume)

Solving for $V_2$: …

Applying Boyle's law ($P_1V_1=P_2V_2$) at constant temperature gives a required pressure of 2.5 bar.

Given

• $P_1 = 1$ bar, $V_1 = 500\ dm^3$
• $V_2 = 200\ dm^3$, same temperature (30°C) throughout, so temperature is constant.

Step 1 — recognize this as an isothermal (constant-T) process

Since temperature doesn't change, Boyle's law (pressure and volume are inversely proportional at constant temperature and amount of gas) applies:

$P_1V_1 = P_2V_2$

Step 2 — solve for $P_2$

$1\ \text{bar} \times 500\ dm^3 = P_2 \times 200\ dm^3$ …